Consider the nuclear fission reaction ₀^1 n+ ₉₂²³⁵ U ₅₆¹⁴⁴ Ba + ₃₆⁸⁹ Kr +3 ₀^1 n . Assuming all the kinetic energy is carried away by the fast neutrons only and total binding energies of ₉₂²³⁵ U , ₅₆¹⁴⁴ Ba and ₃₆⁸⁹ Kr to be 1800 MeV , 1200 MeV and 780 MeV respectively, the average kinetic energy carried by each fast neutron is (in MeV )

Step-by-step solution

Binding energy ₉₂²³⁵ U , B E₁=1800 MeV Binding energy of ₅₆¹⁴⁴ Ba , B E₂=1200 MeV Binding energy of ₃₆⁸⁹ Kr , B E₃=780 MeV Binding energy of reactants =B E₁=1800 MeV Binding energy of products =B E₂=B E₂+B E₃=1200+780=1980 MeV Average kinetic energy carried by each fast neutron Binding energy of products = - Binding energy of reactants 3 = 1980-1800 3 =60 MeV